As a corollary of the double centralizer theorem,
If $\Delta$ is a central finite dimensional division algebra over a field $F$, and if $K$ is the maximal subfield of $\Delta$, then $$\textrm{dim}_F\Delta=(\textrm{dim}_F K)^2.$$
So, for $\Delta_A$, we have $\textrm{dim}_F \Delta_A = (Ind_A)^2$. For $\Delta_B$ and $\Delta_C$,
note that $Z(B)=B\cap C = Z(C)$.
Thus,
$$\textrm{dim}_F\Delta_B=\textrm{dim}_{B\cap C}\Delta_B\cdot\textrm{dim}_F(B\cap C)=(Ind_B)^2\cdot\textrm{dim}_F(B\cap C)$$
$$\textrm{dim}_F\Delta_C=\textrm{dim}_{B\cap C}\Delta_C\cdot\textrm{dim}_F(B\cap C)=(Ind_C)^2\cdot\textrm{dim}_F(B\cap C).$$

I don't know if we may expect a "simple"(pun unintended) answer. You can take $B={\mathbb H}$ the hamiltonian quaternions, embed it in $M_4({\mathbb R})$ and the latter can be embedded in $M_n({\mathbb R})$ any old how. (Then $C$ may be a bit more complicated).

I don't understand you would further embed $M_4(\mathbb{R})$ in $M_n(\mathbb{R})$. Based on OP, the field $\mathbb{F}=\mathbb{R}$, and your example is $B=\mathbb{H}$, $A=M_4(\mathbb{R})$, $C$ is the centralizer of $\mathbb{H}$ in $M_4(\mathbb{R}$. By direct computation, $C$ is isomorphic to $\mathbb{H}$ with different matrices corresponding to $i$, $j$, and $k$.
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i707107Feb 26 '13 at 21:27